Question: You have found the following ages (in years) of 6 bears. The bears are randomly selected from the 29 bears at your local zoo: $ 18,\enspace 2,\enspace 2,\enspace 23,\enspace 21,\enspace 2$ Based on your sample, what is the average age of the bears? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 29 bears, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{44.89} + {86.49} + {86.49} + {136.89} + {94.09} + {86.49}} {{6 - 1}} $ $ {s^2} = \dfrac{{535.34}}{{5}} = {107.07\text{ years}^2} $ We can estimate that the average bear at the zoo is 11.3 years old. There is a variance of 107.07 years $^2$.